# Hamiltonian for spherically symmetric potential

• Vitor Pimenta
In summary, a particle of mass m moving in a central potential V(r) can be described using Hamilton's equations. By using these equations, a one-dimensional equation for the rate of change of momentum in the radial direction, {\dot p_r}, can be obtained in the form {{\dot p}_r} = - \frac{\partial }{{\partial r}}\left[ {{V_{eff}}\left( r \right)} \right]. The effective potential, {V_{eff}}\left( r \right), is a function of r only and includes the effects of the angular momentum, L, which is a constant of the motion. This effective potential takes into account the apparent "centrifugal" effect
Vitor Pimenta

## Homework Statement

A particle of mass m moves in a "central potential" , V(r), where r denotes the radial displacement of the particle from a fixed origin.
From Hamilton´s equations, obtain a "one-dimensional" equation for $${\dot p_r}$$, in the form $${{\dot p}_r} = - \frac{\partial }{{\partial r}}\left[ {{V_{eff}}\left( r \right)} \right]$$, where $${V_{eff}}\left( r \right)$$ denotes an "effective" potential that is a funcion of r only.

## Homework Equations

Hamiltonian: $$H = \frac{{{p_r}^2}}{{2m}} + \frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right)$$ , where L is the angular momentum with respect to the origin, which is a constant of the motion.

$$\frac{{\partial H}}{{\partial r}} = - {\dot p_r}$$

## The Attempt at a Solution

$$\begin{array}{l} {{\dot p}_r} = - \frac{{\partial H}}{{\partial r}} = - \frac{\partial }{{\partial r}}\left[ {\frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right)} \right]\\ \therefore {V_{eff}}\left( r \right) = \frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right) \end{array}$$

The problem is that it doesn´t make sense to me that the effective potential is different than the normal one ( V(r) ). Besides, the force acting on the particle shouldn´t have a dependence on its angular momentum L.

It seems that $$\frac{{{L^2}}}{{2m{r^2}}}$$ has the form of a potential energy the centripetal force could produce ...

Vitor Pimenta said:

## The Attempt at a Solution

$$\begin{array}{l} {{\dot p}_r} = - \frac{{\partial H}}{{\partial r}} = - \frac{\partial }{{\partial r}}\left[ {\frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right)} \right]\\ \therefore {V_{eff}}\left( r \right) = \frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right) \end{array}$$

The problem is that it doesn´t make sense to me that the effective potential is different than the normal one ( V(r) ). Besides, the force acting on the particle shouldn´t have a dependence on its angular momentum L.

The effective potential is different from the normal potential because there's an apparent "centrifugal" effect on the radial motion. The effective potential should not contain the operator ##L^{2}##, you should replace it with its eigenvalue for orbital quantum number ##l##.

hilbert2, thanks for the reply !

I wonder what reference frame the Hamiltonian is about, since it includes the effect of a "false" force (centrifugal). Also, L^2 is not an operator, but a scalar number (which is a constant of the motion), so what was that about replacing it for a quantum number, since we´re not considering quantum mechanics ?

Vitor Pimenta said:
hilbert2, thanks for the reply !

I wonder what reference frame the Hamiltonian is about, since it includes the effect of a "false" force (centrifugal). Also, L^2 is not an operator, but a scalar number (which is a constant of the motion), so what was that about replacing it for a quantum number, since we´re not considering quantum mechanics ?

It is not about any particular reference frame, you can use whatever coordinates you wish (you may even mix spatial and momentum variables as long as your transformation is canonical).

It is not a fictitious effect, your coordinate system is curvilinear and you should expect the motion in one direction (in this case parametrised through the angular momentum, which is a constant of motion) to affect the motion in the other coordinates. Thus, in a rotating coordinate system, it would be the effect of the centrifugal force, while in a fixed coordinate system it is an effect of moving in curvilinear coordinates. The result is of course going to be the same.

Vitor Pimenta said:
Besides, the force acting on the particle shouldn´t have a dependence on its angular momentum L.
This is the effective potential. It tells you how to consider motion in one direction only, in the fixed frame it tells you (for example) the force needed to keep the body moving on a circular path - this does depend on the angular momentum.

I get it somewhat, but this needs further exploration by me (yay, more fun incoming)
Thanks again for all the help

## 1. What is the Hamiltonian for a spherically symmetric potential?

The Hamiltonian for a spherically symmetric potential is a mathematical expression used to describe the total energy of a particle moving in a potential that is symmetric about a point in three dimensions. It is written as H = T + V, where T represents the kinetic energy of the particle and V represents the potential energy.

## 2. How is the Hamiltonian used in quantum mechanics?

In quantum mechanics, the Hamiltonian is used as an operator to describe the energy of a system. It acts on the wave function of the system to determine the time evolution of the system. In the case of a spherically symmetric potential, the Hamiltonian can be solved to find the energy levels and wave functions of the system.

## 3. What is the significance of a spherically symmetric potential?

A spherically symmetric potential is significant because it represents a physical system that has the same potential energy at any point that is equidistant from the center. This means that the potential energy only depends on the distance from the center, making it easier to solve for the energy levels and wave functions using the Hamiltonian.

## 4. How does the Hamiltonian for a spherically symmetric potential relate to the Schrödinger equation?

The Schrödinger equation is a fundamental equation in quantum mechanics that describes the behavior of a particle in a potential. The Hamiltonian for a spherically symmetric potential is used in the Schrödinger equation to solve for the energy levels and wave functions of the system.

## 5. Can the Hamiltonian for a spherically symmetric potential be applied to real-world systems?

Yes, the Hamiltonian for a spherically symmetric potential can be applied to real-world systems, such as atoms or molecules. These systems can be modeled as particles moving in a spherically symmetric potential, making it a useful tool in understanding their behavior and properties.

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